Question

A cylindrical tube open at both ends has a fundamental frequency \(\mathrm{f}\) in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now........ (A) \(\mathrm{f} / 2\) (B) \(\mathrm{f}\) (C) \(3 \mathrm{f} / 4\) (D) \(2 \mathrm{f}\)

Short answer

The new fundamental frequency of the air column when the tube is half-submerged in water is twice the initial frequency. Therefore, the correct answer is (D) \(2f\).

Step-by-step solution

Formula for the fundamental frequency

For an open cylindrical tube, the fundamental frequency (\(f\)) is given by the formula: \[f = \frac{v}{2L}\] where: \(v\) is the speed of sound in air, \(L\) is the length of the air column.

Calculate the length of the air column when half-submerged in water

When the tube is half-submerged in water, the length of the air column is halved. Let \(L'\) represent the new length of the air column: \[L' = \frac{L}{2}\]

Calculate the new fundamental frequency

Using the formula for the fundamental frequency and the new length of the air column (\(L'\)), we can calculate the new fundamental frequency (\(f'\)): \[f' = \frac{v}{2L'}\]

Substitute the new length in the frequency equation

Replace \(L'\) with \(\frac{L}{2}\) in the equation for the new fundamental frequency: \[f' = \frac{v}{2\left(\frac{L}{2}\right)}\]

Simplify the equation and compare the initial and final frequencies

Now we simplify the equation and compare the new frequency (\(f'\)) to the initial frequency (\(f\)): \[f' = \frac{v}{L}\] Comparing this to the initial frequency equation: \[f = \frac{v}{2L}\] We can clearly see that the new fundamental frequency of the air column is twice the initial frequency. Therefore, the correct answer is (D) \(2f\).

Key concepts

Cylindrical Tube

A cylindrical tube is a fundamental shape used in physics to study sound waves. It has a circular cross-section and a specific length, creating a space through which sound waves can travel. The tube can influence how sound resonates within it. - Tubes can be open or closed at the ends, affecting the sound frequencies they produce. - The length of the tube plays a crucial role in determining the natural frequency of sound that the tube can produce. In physics, the cylindrical tube often serves as a model for understanding how sound waves interact with their environment, making it an essential teaching tool. When such a tube is open at both ends, it acts as a resonant cavity for sound waves, allowing for a specific pattern known as standing waves to form inside.

Open-Ended Tube

An open-ended tube, such as the one used in this problem, is a cylindrical tube with both ends open. This means that the air inside can move freely in and out, allowing sound waves to have nodes and antinodes at the ends. - The wave pattern in an open tube results in nodes forming at the ends, with the maximum displacement (antinodes) occurring within. - Such a configuration makes it easy for multiple harmonics to be established. When the tube is open at both ends, it enhances the resonance effect, supporting the fundamental frequency and its harmonics like the second, third, etc., depending on the length of the tube. This is why only certain frequencies can be amplified in an open-ended cylinder.

Speed of Sound in Air

The speed of sound in air (\(v\)) is a key factor in determining the frequency of sound waves traveling through an air column. It is influenced by - Temperature: Sound travels faster in warmer air because particles move more rapidly.- Pressure and Humidity: Increased pressure and higher humidity can also increase the speed slightly.In most conditions at room temperature, the speed of sound in air is approximately 343 meters per second. This speed allows us to calculate the fundamental frequency of a cylindrical tube using the relationship \[f = \frac{v}{2L}\] for open tubes. By understanding the speed of sound, we can predict how sound will behave in different environmental conditions.

Air Column Length

The length of the air column (\(L\)) inside the cylindrical tube greatly affects the frequency of the sound waves it can produce. When the tube is half-submerged in water, as described in the exercise, the effective length of the air column is reduced.- Initially, the full length of the tube was used for calculation: \(L\).- After submersion, only half the length supports standing waves:\(L' = \frac{L}{2}\).This change in length alters the fundamental frequency. Given the relationship \[f = \frac{v}{2L}\] and after submerging the tube, \[f' = \frac{v}{2L'}\], we see that halving the length of the tube halves the frequency. However, the exercise shows that it's actually the denominator that changes, resulting in the frequency doubling. Understanding this concept helps in experiments and applications involving resonance in tubes.